Vector Algebra — NCERT Class 12 Mathematics Notes, MCQs & Flashcards

Q: Which of the following is a vector quantity?

Velocity of 20 m/s towards north. Velocity requires both magnitude (20 m/s) and direction (towards north); others are scalars lacking direction.

Q: If a point P has coordinates (3, 4, 0), what is the magnitude of its position vector?

5. |OP| = √(3² + 4² + 0²) = √(9 + 16) = √25 = 5.

Q: For direction cosines l, m, n of a vector, which statement is NOT correct?

They are proportional to direction ratios. Direction cosines are NOT proportional to direction ratios; rather, they are calculated from ratios via normalization: l = a/√(a²+b²+c²).

Q: Two vectors **a** and **b** are collinear. Which is necessarily true?

**a** and **b** are parallel to the same line. Collinear vectors are parallel to the same line; they may differ in magnitude or direction, so A and C are not necessary.

Q: A displacement vector goes from point A(1,2,3) to point B(4,6,3). What is its magnitude?

5. AB = (4−1, 6−2, 3−3) = (3, 4, 0); |AB| = √(3² + 4²) = √25 = 5.

Q: Which pair represents equal vectors in a square ABCD?

AB and DC. In square ABCD, AB and DC have the same magnitude and direction (both rightward, equal side length), making them equal vectors.

Q: The direction cosines of a vector are l = 1/2, m = 1/2, and n = ?. Find n.

1/√2. Using l² + m² + n² = 1: (1/2)² + (1/2)² + n² = 1 → 1/4 + 1/4 + n² = 1 → n² = 1/2 → n = ±1/√2; taking positive value gives 1/√2.

Q: A girl travels from A to B (displacement **AB**), then from B to C (displacement **BC**). The net displacement is represented by: (i) **AB** + **BC** = **AC** (triangle law); (ii) This follows from free vector property. Which statement is correct?

Both (i) and (ii) are correct. The triangle law (i) is fundamental to vector addition, and (ii) enables parallel shifting so terminal point of AB meets initial point of BC.

Q: If direction ratios of a vector are 1, 2, 2, then its direction cosines are:

1/3, 2/3, 2/3. Magnitude of ratio vector = √(1² + 2² + 2²) = √9 = 3; direction cosines = (1/3, 2/3, 2/3) after normalization.

Q: How does the unit vector differ from the direction of the original vector **a**?

Unit vector has magnitude 1 but same direction as **a**. The unit vector **â** = **a**/|**a**| preserves direction of **a** but scales magnitude to exactly 1.

Chapter Notes

Overview

This chapter introduces vectors, quantities that have both magnitude and direction, as distinct from scalars. It covers the basic concepts of vectors, their types, addition, multiplication by a scalar, and the two key products: scalar (dot) product and vector (cross) product. The central idea is that vectors provide a mathematical framework to represent and manipulate physical quantities like displacement, velocity, and force, enabling geometric and algebraic problem-solving in two and three dimensions.

Key Concepts (24)

Vector

A quantity that has magnitude as well as direction.

e.g. Force applied to push a box: it has a magnitude (e.g., 10 N) and a direction (e.g., east).

Scalar

A quantity that has only magnitude and no direction.

e.g. Your height is 1.6 meters; it is just a number with a unit.

Directed line segment

A line segment with a prescribed direction, having magnitude (length) and direction.

e.g. An arrow from point A to point B on a map showing a displacement of 5 km north.

Initial point

The point from which a vector starts.

e.g. In vector AB, point A is the initial point.

Terminal point

The point where a vector ends.

e.g. In vector AB, point B is the terminal point.

Magnitude of a vector

The length or distance between the initial and terminal points of a vector, denoted as |a| or a.

e.g. For vector AB with A(1,2) and B(4,6), magnitude = √((4-1)²+(6-2)²) = 5 units.

Position vector

The vector representing the position of a point P with respect to the origin O, denoted as OP or r.

e.g. For point P(3,4,5), its position vector is 3i + 4j + 5k.

Direction cosines

The cosines of the angles α, β, γ that a vector makes with the positive x, y, and z axes respectively, denoted by l, m, n.

e.g. For vector i + j + k, direction cosines are (1/√3, 1/√3, 1/√3).

Direction ratios

Numbers proportional to the direction cosines of a vector, denoted as a, b, c.

e.g. For vector 2i + 3j + 4k, direction ratios are 2, 3, 4.

Zero vector (null vector)

A vector whose initial and terminal points coincide, having zero magnitude and no definite direction.

e.g. The vector from a point to itself, e.g., AA.

Unit vector

A vector whose magnitude is unity (1 unit).

e.g. The unit vector in the direction of a is denoted as â = a/|a|.

Coinitial vectors

Two or more vectors having the same initial point.

e.g. Vectors from the same origin O to different points A, B, C.

Collinear vectors

Two or more vectors that are parallel to the same line, irrespective of their magnitudes and directions.

e.g. Vectors 2i + 4j and 3i + 6j are collinear because one is a scalar multiple of the other.

Equal vectors

Two vectors that have the same magnitude and direction, regardless of the positions of their initial points.

e.g. Vector from (0,0) to (1,2) and vector from (3,4) to (4,6) are equal.

Negative of a vector

A vector having the same magnitude as the given vector but opposite direction.

e.g. The negative of vector AB is vector BA, i.e., -AB.

Free vectors

Vectors that can be displaced parallel to themselves without changing their magnitude and direction.

e.g. In this chapter, all vectors are considered free vectors.

Triangle law of vector addition

If two vectors are represented by two sides of a triangle taken in order, then their sum is represented by the third side taken in the opposite order.

e.g. If a girl moves from A to B and then B to C, net displacement AC = AB + BC.

Parallelogram law of vector addition

If two vectors are represented by the two adjacent sides of a parallelogram, then their sum is represented by the diagonal through their common point.

e.g. The resultant velocity of a boat is the diagonal of the parallelogram formed by its engine velocity and river flow velocity.

Scalar (dot) product

The scalar product of two nonzero vectors a and b is defined as a·b = |a||b|cosθ, where θ is the angle between them.

e.g. Work done W = F·s, where F is force and s is displacement.

Vector (cross) product

The vector product of two nonzero vectors a and b is defined as a×b = |a||b|sinθ n̂, where n̂ is a unit vector perpendicular to both a and b, forming a right-handed system.

e.g. Torque τ = r×F, where r is position vector and F is force.

Projection of a vector on a line

The projection of vector a on a directed line l is a vector p with magnitude |a|cosθ, where θ is the angle between a and l.

e.g. The projection of a force vector along a displacement direction gives the component of force doing work.

Component form of a vector

A vector expressed as r = xi + yj + zk, where x, y, z are scalar components along x, y, z axes.

e.g. Position vector of P(2,3,4) is 2i + 3j + 4k.

Section formula (internal division)

The position vector of point R dividing PQ internally in ratio m:n is (m b + n a)/(m+n).

e.g. If P and Q have position vectors a and b, and R divides PQ in ratio 2:1 internally, then OR = (2b + a)/3.

Section formula (external division)

The position vector of point R dividing PQ externally in ratio m:n is (m b - n a)/(m-n).

e.g. If P and Q have position vectors a and b, and R divides PQ externally in ratio 2:1, then OR = (2b - a)/1.

Formulas & Equations (13)

|r| = √(x² + y² + z²)

Reads as: Magnitude of vector r equals square root of x squared plus y squared plus z squared

Variables: r = position vector; x, y, z = scalar components along x, y, z axes (units: same as vector length)

Use when: Finding the length of a vector given in component form

⚠ Forgetting to take the square root after summing the squares

l² + m² + n² = 1

Reads as: Sum of squares of direction cosines equals one

Variables: l, m, n = direction cosines (cosα, cosβ, cosγ) (dimensionless)

Use when: Verifying if given numbers are valid direction cosines

⚠ Applying this to direction ratios (a²+b²+c² ≠ 1 in general)

a·b = |a||b|cosθ

Reads as: a dot b equals magnitude of a times magnitude of b times cosine of theta

Variables: a, b = vectors; |a|, |b| = magnitudes; θ = angle between a and b (radians or degrees)

Use when: Finding the dot product, angle between vectors, or checking perpendicularity

⚠ Using sinθ instead of cosθ

a·b = a₁b₁ + a₂b₂ + a₃b₃

Reads as: a dot b equals a1 b1 plus a2 b2 plus a3 b3

Variables: a₁, a₂, a₃ = components of a; b₁, b₂, b₃ = components of b (same units as vectors)

Use when: Computing dot product when vectors are in component form

⚠ Adding cross terms like a₁b₂

|a×b| = |a||b|sinθ

Reads as: Magnitude of a cross b equals magnitude of a times magnitude of b times sine of theta

Variables: a, b = vectors; |a|, |b| = magnitudes; θ = angle between a and b

Use when: Finding the magnitude of cross product, area of parallelogram or triangle

⚠ Using cosθ instead of sinθ

a×b = (a₂b₃ - a₃b₂)i + (a₃b₁ - a₁b₃)j + (a₁b₂ - a₂b₁)k

Reads as: a cross b equals determinant of i j k with components of a and b

Variables: a₁, a₂, a₃ = components of a; b₁, b₂, b₃ = components of b

Use when: Computing cross product in component form using determinant method

⚠ Sign errors in the j-component (remember the minus sign)

cosθ = (a·b)/(|a||b|)

Reads as: Cosine theta equals a dot b divided by product of magnitudes

Variables: θ = angle between vectors a and b

Use when: Finding the angle between two vectors

⚠ Forgetting to take the inverse cosine after computing the ratio

Projection of a on b = (a·b)/|b|

Reads as: Projection of a on b equals a dot b divided by magnitude of b

Variables: a = vector being projected; b = vector on which projection is taken

Use when: Finding the scalar projection of one vector onto another

⚠ Dividing by |a| instead of |b|

Position vector of R (internal division) = (m b + n a)/(m+n)

Reads as: Position vector of R equals m times b plus n times a divided by m plus n

Variables: a, b = position vectors of P and Q; m:n = ratio in which R divides PQ internally

Use when: Finding the coordinates of a point dividing a line segment internally in a given ratio

⚠ Swapping m and n in the numerator

Position vector of R (external division) = (m b - n a)/(m-n)

Reads as: Position vector of R equals m times b minus n times a divided by m minus n

Variables: a, b = position vectors of P and Q; m:n = ratio in which R divides PQ externally

Use when: Finding the coordinates of a point dividing a line segment externally in a given ratio

⚠ Using plus sign instead of minus in numerator

Vector joining P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) = (x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k

Reads as: Vector from P to Q equals x2 minus x1 i plus y2 minus y1 j plus z2 minus z1 k

Variables: P, Q = points with coordinates (x₁,y₁,z₁) and (x₂,y₂,z₂)

Use when: Finding the vector from one point to another

⚠ Subtracting in the wrong order (terminal minus initial)

|a+b| ≤ |a| + |b| (Triangle inequality)

Reads as: Magnitude of a plus b is less than or equal to magnitude of a plus magnitude of b

Variables: a, b = vectors

Use when: Proving inequalities or checking if three points are collinear (equality holds)

⚠ Assuming equality always holds

|a·b| ≤ |a||b| (Cauchy-Schwarz inequality)

Reads as: Absolute value of a dot b is less than or equal to product of magnitudes

Variables: a, b = vectors

Use when: Proving inequalities or bounding dot product values

⚠ Forgetting the absolute value on the left side

How It Works

Triangle Law of Vector Addition

Place the vectors such that the initial point of the second vector coincides with the terminal point of the first vector.
Draw the resultant vector from the initial point of the first vector to the terminal point of the second vector.
The resultant vector represents the sum of the two vectors.

💡 The sides of the triangle must be taken in order; the resultant is the third side in the opposite direction.

Parallelogram Law of Vector Addition

Place the two vectors with a common initial point.
Complete the parallelogram by drawing lines parallel to each vector from the terminal points.
The diagonal through the common initial point represents the sum of the two vectors.

💡 This is equivalent to the triangle law; the diagonal is the resultant.

Multiplication of a Vector by a Scalar

Identify the scalar λ and the vector a.
The new vector λa has magnitude |λ| times the magnitude of a.
The direction of λa is same as a if λ > 0, opposite if λ < 0.

💡 λ = 0 gives the zero vector; λ = -1 gives the negative of the vector.

Finding Unit Vector in the Direction of a Given Vector

Compute the magnitude |a| of the given vector a.
Divide each component of a by |a|.
The resulting vector â = a/|a| is the unit vector in the direction of a.

💡 Unit vector has magnitude exactly 1; it is dimensionless.

Computing Dot Product in Component Form

Write both vectors in component form: a = a₁i + a₂j + a₃k, b = b₁i + b₂j + b₃k.
Multiply corresponding components: a₁b₁, a₂b₂, a₃b₃.
Sum these products: a·b = a₁b₁ + a₂b₂ + a₃b₃.

💡 Dot product is a scalar, not a vector. Do not multiply cross components.

Computing Cross Product Using Determinant

Write the vectors in component form: a = a₁i + a₂j + a₃k, b = b₁i + b₂j + b₃k.
Set up a 3x3 determinant with i, j, k in the first row, components of a in the second, components of b in the third.
Evaluate the determinant: i(a₂b₃ - a₃b₂) - j(a₁b₃ - a₃b₁) + k(a₁b₂ - a₂b₁).

💡 The j-component has a minus sign; the result is a vector perpendicular to both a and b.

Finding Projection of a Vector on Another Vector

Compute the dot product a·b.
Compute the magnitude |b| of the vector on which projection is taken.
The scalar projection is (a·b)/|b|.
The vector projection is ((a·b)/|b|²) b.

💡 Projection is a scalar; the vector projection is a vector along b.

Section Formula for Internal Division

Identify the position vectors a and b of points P and Q.
Identify the ratio m:n in which R divides PQ internally.
Apply formula: OR = (m b + n a)/(m+n).

💡 The point closer to P has the smaller ratio; m corresponds to Q, n to P.

Section Formula for External Division

Identify the position vectors a and b of points P and Q.
Identify the ratio m:n in which R divides PQ externally.
Apply formula: OR = (m b - n a)/(m-n).

💡 The denominator is m-n, not m+n; the numerator uses subtraction.

Exam Traps — Don't Make These Mistakes

⚠ Confusing direction cosines with direction ratios

✗ Direction cosines are a, b, c such that a² + b² + c² = 1.

→ Direction cosines are l, m, n (cosα, cosβ, cosγ) and satisfy l²+m²+n²=1. Direction ratios are a, b, c which are proportional to l, m, n, and a²+b²+c² ≠ 1 in general.

✓ Direction cosines l, m, n satisfy l² + m² + n² = 1. Direction ratios a, b, c are proportional to l, m, n, and a² + b² + c² ≠ 1 in general.

⚠ Sign error in cross product j-component

✗ a×b = (a₂b₃ - a₃b₂)i + (a₁b₃ - a₃b₁)j + (a₁b₂ - a₂b₁)k

→ The j-component should have a minus sign: - (a₁b₃ - a₃b₁) or (a₃b₁ - a₁b₃).

✓ a×b = (a₂b₃ - a₃b₂)i - (a₁b₃ - a₃b₁)j + (a₁b₂ - a₂b₁)k

⚠ Using triangle inequality incorrectly for collinearity

✗ If |a+b| = |a| + |b|, then a and b are collinear.

→ Equality holds only when a and b are in the same direction (θ=0), not when they are opposite (θ=π gives |a+b| = ||a|-|b||).

✓ If |a+b| = |a| + |b|, then a and b are parallel and in the same direction (θ=0).

⚠ Forgetting that dot product of perpendicular vectors is zero

✗ If a·b = 0, then a and b are perpendicular.

→ This is true only if both a and b are nonzero vectors. If either is zero, the dot product is zero but the vectors are not necessarily perpendicular.

✓ If a and b are nonzero vectors and a·b = 0, then a and b are perpendicular.

⚠ Mistaking projection of a on b for projection of b on a

✗ Projection of a on b = (a·b)/|a|

→ The projection of a on b is (a·b)/|b|, not divided by |a|. The denominator is the magnitude of the vector on which projection is taken.

✓ Projection of a on b = (a·b)/|b|

Compare & Contrast

Result type

Scalar (dot) product

Vector (cross) product

Dot product yields a scalar (real number); cross product yields a vector.

Commutativity

Scalar (dot) product

Vector (cross) product

Dot product is commutative (a·b = b·a); cross product is anti-commutative (a×b = - b×a).

Definition

Direction cosines

Direction ratios

Direction cosines are the cosines of the angles made with axes (l, m, n) and satisfy l²+m²+n²=1; direction ratios are any numbers proportional to direction cosines (a, b, c) and a²+b²+c² ≠ 1 in general.

Formula

Internal division

External division

Internal division uses OR = (m b + n a)/(m+n); external division uses OR = (m b - n a)/(m-n).

Exam Focus

Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks), 1 LA (5 marks) — total 10 marks typically

Most asked: Finding angle between two vectors using dot product · Finding unit vector perpendicular to two vectors using cross product · Section formula (internal and external division) · Projection of a vector on another vector · Collinearity and perpendicularity conditions · Area of triangle/parallelogram using cross product

Don't skip: Long answer (5 marks) on cross product: finding area of triangle or parallelogram, or proving vectors form a right-angled triangle; Short answer (3 marks) on section formula or projection

One-Line Revision (read aloud before the exam)

A vector has both magnitude and direction; a scalar has only magnitude.
The magnitude of a vector r = xi + yj + zk is |r| = √(x² + y² + z²).
Direction cosines l, m, n satisfy l² + m² + n² = 1.
Triangle law: AB + BC = AC.
Parallelogram law: diagonal through common point is the sum.
Dot product a·b = |a||b|cosθ = a₁b₁ + a₂b₂ + a₃b₃.
Cross product a×b = |a||b|sinθ n̂, with direction given by right-hand rule.
Two nonzero vectors are perpendicular if a·b = 0.
Two nonzero vectors are parallel (collinear) if a×b = 0.
Unit vector in direction of a is â = a/|a|.
Projection of a on b = (a·b)/|b|.
Area of parallelogram = |a×b|; area of triangle = ½|a×b|.
Section formula (internal): OR = (m b + n a)/(m+n).
Section formula (external): OR = (m b - n a)/(m-n).
Vector joining P(x₁,y₁,z₁) to Q(x₂,y₂,z₂) is (x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k.
Triangle inequality: |a+b| ≤ |a| + |b|.
Cauchy-Schwarz inequality: |a·b| ≤ |a||b|.
Zero vector has zero magnitude and no definite direction.
Cross product is anti-commutative: a×b = - b×a.
For unit vectors i, j, k: i×j = k, j×k = i, k×i = j.

MCQs — 10 Questions with Answers

Q1. Which of the following is a vector quantity?

A. Speed of 20 m/s
B. Velocity of 20 m/s towards north ✓
C. Temperature of 30°C
D. Mass of 5 kg

Answer: B — Velocity requires both magnitude (20 m/s) and direction (towards north); others are scalars lacking direction.

Q2. If a point P has coordinates (3, 4, 0), what is the magnitude of its position vector?

A. 5 ✓
B. 7
C. √7
D. 12

Answer: A — |OP| = √(3² + 4² + 0²) = √(9 + 16) = √25 = 5.

Q3. For direction cosines l, m, n of a vector, which statement is NOT correct?

A. l² + m² + n² = 1
B. Each lies between −1 and 1
C. They are unique for any non-zero vector
D. They are proportional to direction ratios ✓

Answer: D — Direction cosines are NOT proportional to direction ratios; rather, they are calculated from ratios via normalization: l = a/√(a²+b²+c²).

Q4. Two vectors **a** and **b** are collinear. Which is necessarily true?

A. |**a**| = |**b**|
B. **a** and **b** are parallel to the same line ✓
C. **a** = **b**
D. Both have the same initial point

Answer: B — Collinear vectors are parallel to the same line; they may differ in magnitude or direction, so A and C are not necessary.

Q5. A displacement vector goes from point A(1,2,3) to point B(4,6,3). What is its magnitude?

A. 5 ✓
B. √41
C. 7
D. √34

Answer: A — AB = (4−1, 6−2, 3−3) = (3, 4, 0); |AB| = √(3² + 4²) = √25 = 5.

Q6. Which pair represents equal vectors in a square ABCD?

A. AB and CD
B. AB and BC
C. AB and DC ✓
D. AD and BC

Answer: C — In square ABCD, AB and DC have the same magnitude and direction (both rightward, equal side length), making them equal vectors.

Q7. The direction cosines of a vector are l = 1/2, m = 1/2, and n = ?. Find n.

A. 1/2
B. 1/√2 ✓
C. √2/2
D. 0

Answer: B — Using l² + m² + n² = 1: (1/2)² + (1/2)² + n² = 1 → 1/4 + 1/4 + n² = 1 → n² = 1/2 → n = ±1/√2; taking positive value gives 1/√2.

Q8. A girl travels from A to B (displacement **AB**), then from B to C (displacement **BC**). The net displacement is represented by: (i) **AB** + **BC** = **AC** (triangle law); (ii) This follows from free vector property. Which statement is correct?

A. Both (i) and (ii) are correct ✓
B. Only (i) is correct
C. Only (ii) is correct
D. Neither is correct

Answer: A — The triangle law (i) is fundamental to vector addition, and (ii) enables parallel shifting so terminal point of AB meets initial point of BC.

Q9. If direction ratios of a vector are 1, 2, 2, then its direction cosines are:

A. 1/3, 2/3, 2/3 ✓
B. 1, 2, 2
C. 1/5, 2/5, 2/5
D. Cannot be determined

Answer: A — Magnitude of ratio vector = √(1² + 2² + 2²) = √9 = 3; direction cosines = (1/3, 2/3, 2/3) after normalization.

Q10. How does the unit vector differ from the direction of the original vector **a**?

A. Unit vector points in opposite direction
B. Unit vector has magnitude 1 but same direction as **a** ✓
C. Unit vector cannot be defined for negative vectors
D. Direction is completely different from **a**

Answer: B — The unit vector **â** = **a**/|**a**| preserves direction of **a** but scales magnitude to exactly 1.

Flashcards

What is a vector? Define with one example from physics.

A vector is a quantity with both magnitude and direction (e.g., displacement 40 km north); a directed line segment from A to B represents it.

What is the position vector of point P(x, y, z)?

The vector OP from origin O(0,0,0) to point P(x,y,z), with magnitude |OP| = √(x² + y² + z²).

Define direction cosines and state the fundamental identity.

Direction cosines l, m, n are cosα, cosβ, cosγ (cosines of angles with x, y, z axes); key identity: l² + m² + n² = 1.

What is the relationship between direction cosines and coordinates?

If |OP| = r, then coordinates of P are (lr, mr, nr) where l = x/r, m = y/r, n = z/r.

Distinguish between direction cosines and direction ratios.

Direction cosines are l, m, n with l² + m² + n² = 1; direction ratios a, b, c are proportional to them with a² + b² + c² ≠ 1 in general.

What is a unit vector and how is it denoted?

A unit vector has magnitude 1; the unit vector in direction of vector a is denoted â = a/|a|.

State the triangle law of vector addition with a diagram example.

If a girl moves A→B then B→C, net displacement is AC = AB + BC; vectors add by placing terminal point of first at initial point of second.

What are collinear vectors and how do they differ from equal vectors?

Collinear vectors are parallel to the same line (may differ in magnitude/direction); equal vectors have identical magnitude AND direction regardless of position.

Define the negative of a vector with notation.

The negative of vector AB is the vector BA (same magnitude, opposite direction), denoted −AB = BA.

What are coinitial vectors and give an example from a diagram.

Two or more vectors sharing the same initial (starting) point are coinitial; example: vectors AB and AD from same point A in a square.

Important Board Questions

Classify the following as scalar or vector quantities: (i) 10 kg, (ii) 5 m/s north, (iii) 100° angle, (iv) Force of 50 Newton. Give reason for each classification. [2 marks]

Scalars have magnitude only (e.g., mass, temperature); vectors require both magnitude and direction (e.g., velocity, force). Check if each quantity specifies a direction.

A point P has coordinates (1, 2, 2) in 3D space. Find: (i) its position vector OP, (ii) magnitude |OP|, (iii) direction cosines l, m, n. Show all working steps. [5 marks]

Position vector **OP** = (1, 2, 2); find magnitude using √(x² + y² + z²); then direction cosines l = x/|OP|, m = y/|OP|, n = z/|OP|. Verify: l² + m² + n² = 1.

Given direction ratios of a vector are 2, −2, 1. Find its direction cosines. Also, if the vector has magnitude 18, find the vector's components. Derive the relation between direction cosines and direction ratios fully. [6 marks]

First normalize direction ratios: √(2² + (−2)² + 1²) = 3. Then direction cosines = (2/3, −2/3, 1/3). Components = direction cosines × magnitude = 18 × (2/3, −2/3, 1/3) = (12, −12, 6). Show that direction cosines satisfy l² + m² + n² = 1 as proof.

Assertion-Reason Questions for Vector Algebra

Q1. Assertion (A):

The magnitude of a position vector **r** = x**i** + y**j** + z**k** is always equal to √(x² + y² + z²).

Reason (R):

The distance formula in three-dimensional space is derived from the Pythagorean theorem applied successively in three perpendicular directions.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why the magnitude formula takes this specific form. The distance from origin to point P(x,y,z) is indeed √(x² + y² + z²), which comes from applying Pythagorean theorem three times.

Q2. Assertion (A):

Two collinear vectors must have the same magnitude.

Reason (R):

Collinear vectors are parallel to the same line and can have any magnitude as long as they maintain the same or opposite direction.

Answer: D — A is false but R is true — The assertion is false. From Section 10.3, collinear vectors are those parallel to the same line regardless of magnitudes and directions. Two collinear vectors can have different magnitudes. The reason is true but doesn't explain the false assertion correctly.

Q3. Assertion (A):

If direction cosines of a vector are l, m, n, then the direction ratios are al, am, an for any non-zero constant a.

Reason (R):

Direction ratios are proportional to direction cosines, and multiplying by any non-zero constant preserves the proportionality relationship.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both are true. Direction ratios (a, b, c) satisfy the relationship a:b:c = l:m:n, so they can be written as (kl, km, kn) for some constant k. The reason explains why this proportionality is valid.

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Case-Based Questions (CBQ) for Vector Algebra

Case Study:

A surveyor needs to map a triangular plot of land ABC. She measures the displacement from A to B as 100 m due east, and then from B to C as 60 m at an angle of 60° north of east. The surveyor uses vector addition to find the net displacement from A to C. She also needs to calculate the direction cosines of the vector **AC** to determine its orientation with respect to the coordinate axes.

Q1. The components of displacement **AB** in a coordinate system where east is the positive x-axis and north is the positive y-axis are:

A. (100, 0) m ✓
B. (0, 100) m
C. (100, 100) m
D. (50, 50√3) m

Answer: A

Q2. The x and y components of displacement **BC** (60 m at 60° north of east) are approximately:

A. (30, 30√3) m
B. (60, 0) m
C. (30√3, 30) m
D. (51.96, 30) m ✓

Answer: D

Q3. The magnitude of net displacement **AC** is approximately:

A. 160 m
B. 152 m
C. 152.97 m ✓
D. 140 m

Answer: C

Q4. The direction of **AC** from the positive x-axis is approximately:

A. 30°
B. 45°
C. 22.1° ✓
D. 60°

Answer: C

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Previous Year Questions (PYQ) for Vector Algebra

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

The value of p for which vectors î + 2ĵ + 3k̂ and 2î – pĵ + k̂ are perpendicular to each other is

0
1
5/2
–5/2

CBSE 2026 · Set 65/1/1 · 1 mark

The value of m for which the points with position vectors –î – ĵ + 2k̂, 2î + mĵ + 5k̂ and 3î + 11ĵ + 6k̂ are collinear, is

8
–8
2
5/2

CBSE 2026 · Set 65/1/1 · 1 mark

If |a| = 8, |b| = 3 and |a × b| = 12, then the value of |a · b| is

6√3
8√3
12√3
√(3·12)

CBSE 2026 · Set 65/1/1 · 2 marks

Find the vector of magnitude 14 in the direction of QP, where P and Q are the points (1, 3, 2) and (–1, 0, 8) respectively.

CBSE 2026 · Set 65/1/1 · 2 marks

Vectors a = 3î – 2ĵ + 2k̂ and b = î + 2k̂ represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths.

CBSE 2026 · Set 65/1/2 · 2 marks

A vector a of magnitude 14 has direction ratios <2, 3, –6>. Find the projection of the vector a on î.

CBSE 2026 · Set 65/1/3 · 1 mark

The value of p for which vectors i + 2j + 3k and 2i – pj + k are perpendicular to each other is

0
1
5/2
-5/2

CBSE 2026 · Set 65/1/3 · 2 marks

Using vectors, find the area of triangle ABC with vertices A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1).

CBSE 2026 · Set 65/1/3 · 2 marks

Vectors a = 3i – 2j + 2k and b = i + 2k represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths.

CBSE 2026 · Set 65/2/1 · 1 mark

If |a| = 5 and a·b = -2 and |a·b| = 1, then the sum of greatest and the smallest value of |a| is

-5
5
10
15

CBSE 2026 · Set 65/2/1 · 1 mark

Vector of magnitude 3 making equal angles with x and y axes and perpendicular to z axis is

î + 2√2ĵ
3k̂
(3√2/2)î + (3√2/2)ĵ
3î + 3ĵ + 3k̂

CBSE 2026 · Set 65/2/1 · 1 mark

Assertion (A): For two vectors a and b, |a × b|² + (a·b)² = |a|²|b|². Reason (R): |a × b| = (a·b) tan θ, where 0 ≤ θ ≤ π/2.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.

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Class 12 Maths Formula Sheet

Every formula from all 13 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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